Find the mass of AlCl3 that is produced when 25.0 grams of Al2O3 react with excess HCl according to the following equation.

Al2O3(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2O(l)
Group of answer choices

72.9 g

155 g

65.4 g

16.3 g

32.6 g

The correct answer to the question is 65.4 g

The balanced equation for the reaction is given below

Next, we shall determine the mass of Al₂O₃ that reacted and the mass of AlCl₃ produced from the balanced.

Molar mass of Al₂O₃ = (27×2) + (3×16)

= 54 + 48

= 102 g/mol

Mass of Al₂O₃ from the balanced equation = 1 × 102 = 102 g

Molar mass of AlCl₃ = 27 + (35.5×3)

= 27 + 106.5

= 133.5 g/mol

Mass of AlCl₃ from the balanced equation = 2 × 133.5 = 267 g

Thus,

From the balanced equation above,

Finally, we shall determine the mass of AlCl₃ produced by the reaction of 35 g of Al₂O₃. This is illustrated below:

From the balanced equation above,

102 g of Al₂O₃ reacted to produce 267 g of AlCl₃.

Therefore,

25 g of Al₂O₃ will react to produce = = 65.4 g of AlCl₃.

Thus, 65.4 g of AlCl₃ were obtained from the reaction.

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Mass = 65.4 g

Explanation:

Given data:

Mass of Al₂O₃ = 25 g

Mass of AlCl₃ = ?

Solution:

Chemical equation:

Al₂O₃ + 6HCl  →   2AlCl₃ + 3H₂O

Number of moles of Al₂O₃:

Number of moles = mass/ molar mass

Number of moles = 25 g/  101.96 g/mol

Number of moles = 0.2452 mol

Now we will compare the moles of Al₂O₃  and AlCl₃.

                          Al₂O₃         :            AlCl₃

                              1             :               2

                             0.2452    :          2×0.2452 = 0.4904 mol

Mass of AlCl₃:

Mass = number of moles × molar mass

Mass = 0.4904 mol × 133.34 g/mol

Mass = 65.4 g

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.